a 

oki

Xét 2017 /2018 và 2018/2019

1-2017/2018=1/2018

1-2018/2019=1/2019

mà 1/2018>1/2019=>2017/2018<2018/2019

Tương tự có:2020/2019>2021/2020

=>2017/2018+2010/2019<2018/2019+2021/2020

Tham khảo:

\(\dfrac{-2019}{2019}=-1\)

\(\dfrac{-2021}{2020}=-1,004\)

\(\Rightarrow\dfrac{-2019}{2019}>\dfrac{-2021}{2020}\)

\(-\dfrac{2019}{2019}=-1\)

\(\dfrac{-2021}{2020}< -1\)

Do đó: \(-\dfrac{2019}{2019}>\dfrac{-2021}{2020}\)

Ta có 

A = 2017/2019 =1 - 2/2019

B = 2021/2023 = 1 - 2/2013

MÀ 2/2019 < 2/2013 => 1 - 2/2019 > 1 - 2/2013 hay A > B

Vậy A > B

Easy mà bạn : 

Ta có : 

\(A=\frac{2017}{2019}=1-\frac{2}{2019}\)

\(B=\frac{2021}{2023}=1-\frac{2}{2023}\)

Do \(\frac{2}{2019}>\frac{2}{2023}\)

\(\Rightarrow1-\frac{2}{2019}< 1-\frac{2}{2023}\)

\(\Rightarrow A< B\)

~ 

Ta có : \(1-\frac{2017}{2019}=\frac{2}{2019};1-\frac{2021}{2023}=\frac{2}{2023}\)

Vì \(\frac{2}{2019}>\frac{2}{2023}\)nên \(\frac{2017}{2019}< \frac{2021}{2023}\).

~~~~~~~~~~~HOK TỐT~~~~~~~~~~~~

Vì 2019 + 2020 < 2019 + 2021 nên A < B

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)

\(\dfrac{2021}{2019}và\dfrac{2023}{2021}\)

\(\Rightarrow\dfrac{2021}{2019}-\dfrac{2}{2019}=\dfrac{2023}{2021}-\dfrac{2}{2021}\left(=1\right)\)

\(\Rightarrow\dfrac{2}{2019}>\dfrac{2}{2021}\Rightarrow\dfrac{2021}{2019}< \dfrac{2023}{2021}\)

Chứng minh bđt phụ nếu a>b \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(vớim\in N^{\circledast}\right)\Rightarrow a\left(b+m\right)>b\left(a+m\right)\Rightarrow ab+am>ab+bm\Rightarrow am>bm\Rightarrow a>b\) \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(1\right)\)

Áp dụng bđt (1) có :

\(2021>2019\Rightarrow\dfrac{2021}{2019}>\dfrac{2021+2}{2019+2}=\dfrac{2023}{2021}\)